素數C ++程序

Question

我不確定是否應該在這里詢問程序員，還是我應該問程序員，但是我一直在試圖弄清為什么該程序無法運行，盡管我發現了一些錯誤，但即使返回x也不是素數。

#include <iostream>
using namespace std;


  bool primetest(int a) {
 int i;
 //Halve the user input to find where to stop dividing to (it will remove decimal point as it is an integer)
 int b = a / 2;
 //Loop through, for each division to test if it has a factor (it starts at 2, as 1 will always divide)
 for (i = 2; i < b; i++) {
     //If  the user input has no remainder then it cannot be a prime and the loop can stop (break)
     if (a % i == 0) {
           return(0);
           break;
     }
     //Other wise if the user input does have a remainder and is the last of the loop, return true (it is a prime)
              else if ((a % i != 0) && (i == a -1)) {
          return (1);
          break;
     }
 }   
}

 int main(void) {
int user;
cout << "Enter a number to test if it is a prime or not: ";
cin >> user;
if (primetest(user)) {
                   cout << user << " is a prime number.";
}
else  {
      cout << user<< " is not a prime number.";
}
cout << "\n\nPress enter to exit...";
getchar();
getchar();
return 0;
}

抱歉，如果它太本地化（在這種情況下，您可以建議我在哪里問這樣的具體問題？）

我應該補充一點，我對C ++（和一般的編程）非常陌生

這僅僅是為了測試功能和控件。

Answer 1

i永遠不會等於a - 1你只會上升到b - 1 。 b為a/2 ，永遠不會導致匹配。

這意味着將返回1的循環結束條件永遠不會成立。

如果是質數，則循環結束。 這會導致不確定的行為，因為那里沒有return語句。 Clang發出了警告，沒有任何特殊標志：

example.cpp:22:1: warning: control may reach end of non-void function
      [-Wreturn-type]
}
^
1 warning generated.

如果您的編譯器沒有警告您，則需要打開更多警告標志。 例如，添加-Wall會在使用GCC時發出警告：

example.cpp: In function ‘bool primetest(int)’:
example.cpp:22: warning: control reaches end of non-void function

總體而言，素數檢查循環要復雜得多。 假設你只關心的值a大於或等於2 ：

bool primetest(int a)
{
    int b = sqrt(a); // only need to test up to the square root of the input

    for (int i = 2; i <= b; i++)
    {
        if (a % i == 0)
           return false;
   }

   // if the loop completed, a is prime
   return true;
}

如果要處理所有int值，則可以添加if (a < 2) return false; 一開始。

Answer 2

您的邏輯不正確。 您正在使用此表達式(i == a -1)) ，就像卡爾所說的那樣，它永遠不可能是正確的。

例如：-

 If a = 11

 b = a/2 = 5  (Fractional part truncated)

因此，您正在循環運行直到i<5 。 所以i永遠不能等於a-1因為在這種情況下i的最大值將是4而a-1值將是10

Answer 3

您只需檢查直到平方根即可。 但是下面是對代碼的一些修改，以使其正常工作。

#include <iostream>
using namespace std;
bool primetest(int a) {
int i;
//Halve the user input to find where to stop dividing to (it will remove decimal point as it is an integer)
int b = a / 2;
//Loop through, for each division to test if it has a factor (it starts at 2, as 1 will always divide)
for (i = 2; i <= b; i++) {
 //If  the user input has no remainder then it cannot be a prime and the loop can stop (break)
 if (a % i == 0) {
       return(0);

 }
}
//this return invokes only when it doesn't has factor
return 1;   
}

int main(void) {
  int user;
  cout << "Enter a number to test if it is a prime or not: ";
  cin >> user;
  if (primetest(user)) {
               cout << user << " is a prime number.";
  }
  else  {
     cout << user<< " is not a prime number.";
  }

return 0;

}

Answer 4

看一下這個：

//Prime Numbers generation in C++
//Using for loops and conditional structures
#include <iostream>
using namespace std;

int main()
{
int a = 2;       //start from 2
long long int b = 1000;     //ends at 1000

for (int i = a; i <= b; i++)
{

 for (int j = 2; j <= i; j++)
 {
    if (!(i%j)&&(i!=j))    //Condition for not prime
        {
            break;
        }

    if (j==i)             //condition for Prime Numbers
        {
              cout << i << endl;

        }
 }
}
}

Answer 5

main()
{
    int i,j,x,box;
    for (i=10;i<=99;i++)
    {
        box=0;
        x=i/2;
        for (j=2;j<=x;j++)
            if (i%j==0) box++;
        if (box==0) cout<<i<<" is a prime number";
        else cout<<i<<" is a composite number";
        cout<<"\n";
        getch();
    }
}

Answer 6

這是查找素數直到所有用戶輸入的數字的完整解決方案。

#include <iostream.h>
#include <conio.h>
using namespace std;

main() 
{
 int num, i, countFactors;
 int a;
 cout << "Enter number " << endl;
 cin >> a;

 for (num = 1; num <= a; num++)
 {
  countFactors = 0;
  for (i = 2; i <= num; i++)
  {
   //if a factor exists from 2 up to the number, count Factors
   if (num % i == 0)
   {
    countFactors++;    
   }
  }

  //a prime number has only itself as a factor
  if (countFactors == 1)
  {
   cout << num << ", ";
  }
 }

 getch();
}

Answer 7

一種方法是使用篩選算法，例如Eratosthenes的篩選器 。 這是一種非常有效的快速方法。

bool isPrime(int number){
  if(number == 2 || number == 3 | number == 5 || number == 7) return true;
  return ((number % 2) && (number % 3) && (number % 5) && (number % 7));
}